Question

分解因式1(3x+5)^2(3x+7)(x+1)-4 2x^4-x^3-6x^2-x+2 一定用换元法，谢谢

Answer 1

(3x+5)^2(3x+7)(x+1)-4
=(9x^2+30x+15)(3x^2+10x+7)-4
=3(3x^2+10x+5)^2+6(3x^2+10x+5)-4,①
设y=3x^2+10x+5,则
①=3y^2+6y-4=3[y-(-3+√21)/3][y-(-3-√21)/3]
=3[3x^2+10x+5-(-3+√21)/3][3x^2+10x+5-(-3-√21)/3]
=[9x^2+30x+18-√21][3x^2+10x+5-(-3-√21)/3].
在实数范围内，第一个因式还可以分解.此处从略.
2x^4-x^3-6x^2-x+2
=x^2[2x^2+2/x^2-x-1/x-6],②
设u=x+1/x,
②=x^2[2(u^2-2)-u-6]
=x^2[2u^2-u-10]
=2x^2(u-5/2)(u+2)
=2(x^2-5x/2+1)(x^2+2x+1)
=(2x-1)(x-2)(x+1)^2.